Math 30 - Day 15 - Dividing Polynomials

Overview

Christopher Lee reviews polynomial graphing, then teaches polynomial long division and synthetic division through worked examples, guided practice, and checks using the division algorithm. The lesson concludes with the remainder theorem, showing how dividing by x − k gives the remainder f(k), including a fifth-degree example evaluated at x = −3.

Key takeaways

Polynomial long division follows the same divide, multiply, subtract, and bring-down cycle as whole-number division; its result satisfies dividend = divisor × quotient + remainder.
A polynomial's remainder must be zero or have degree lower than the divisor, and a zero remainder proves that the divisor is a factor.
Synthetic division is a coefficient-only shortcut for divisors of the form x − k; use the root k, not the divisor's written sign, in the calculation.
When a polynomial omits a power, insert a zero coefficient so synthetic division keeps every exponent in the correct position.
For division by x − k, the remainder theorem gives the remainder directly as f(k), turning polynomial evaluation into a synthetic-division calculation.
At an x-intercept, even multiplicity produces a bounce and odd multiplicity produces a crossing; higher multiplicities also flatten the graph near the intercept.

Chapters

0:00 Polynomial Graphing: Intercepts, Symmetry, and End Behavior

The graphing checklist begins with x- and y-intercepts, then checks even or odd symmetry.
Use zero multiplicities to determine whether a graph crosses or bounces at each x-intercept.
Determine end behavior from the leading term and keep turning points to at most one less than the polynomial's degree.

5:00 Finding Intercepts of −2(x + 3)²(x − 5)

Substituting x = 0 gives the y-intercept (0, 90).
Setting each variable factor to zero gives x-intercepts (−3, 0) and (5, 0).
The constant factor −2 does not create an x-intercept.

10:00 Degree and End Behavior of the Cubic Example

The degree of −2(x + 3)²(x − 5) is 3 because its factor degrees add to 2 + 1.
A positive cubic has left-down, right-up end behavior; a negative leading coefficient reverses those directions.
The graph's end behavior follows its odd degree, while the leading coefficient determines which end rises.

15:00 Multiplicity Determines Crossing and Bounce Behavior

At x = −3, multiplicity 2 is even, so the graph touches the x-axis and turns around.
At x = 5, multiplicity 1 is odd, so the graph crosses the x-axis.
Higher multiplicities make the graph flatter near the intercept, as with powers such as 4 or 7.

19:00 Lincoln Memorial Volume Motivates Polynomial Division

The Lincoln Memorial example uses dimensions 61.5 m by 40 m by 30 m, giving a volume of 73,800 m³.
For a rectangular solid, height equals volume divided by length and width.
A polynomial volume, 3x⁴ − 3x³ − 33x² + 54x, is introduced with length 3x and width x − 2.

22:00 Reviewing Long Division with 178 Divided by 3

Long division repeatedly divides, multiplies, subtracts, and brings down the next digit.
For 178 ÷ 3, the quotient is 59 with remainder 1, or 59⅓.
The division algorithm checks the result: 178 = 3 × 59 + 1.

25:00 Polynomial Long Division of a Cubic by x + 2

Divide 2x³ − 3x² + 4x + 5 by x + 2, starting with 2x³ ÷ x = 2x².
Repeated multiply-and-subtract steps produce quotient 2x² − 7x + 18.
The remainder is −31, so the result is 2x² − 7x + 18 − 31/(x + 2).

29:00 Division Algorithm and Long-Division Procedure

For dividend f(x) and nonzero divisor d(x), the division algorithm gives f(x) = d(x)q(x) + r(x).
The remainder is zero or has degree less than the divisor; a zero remainder means the divisor is a factor.
The long-division steps are to divide leading terms, multiply by the divisor, subtract, bring down, and repeat.

32:00 Dividing 5x² + 3x − 2 by x + 1

The leading-term calculation 5x² ÷ x gives the first quotient term, 5x.
Subtracting the product of x + 1 and 5x leaves −2x − 2.
The second quotient term is −2, producing quotient 5x − 2 with remainder zero.

39:00 Guided Practice: Handling Polynomial Subtraction

Unlike terms such as 3x² and −5x cannot be combined because their exponents differ.
Subtraction changes signs through the coefficients, so subtracting a negative term becomes addition.
The class revisits the long-division setup and brings down each next dividend term in sequence.

49:00 Worked Division of 6x³ + 11x² − 31x + 15

Divide 6x³ + 11x² − 31x + 15 by 3x − 2 using leading-term division.
The quotient terms are 2x², 5x, and −7.
The final remainder is 1, giving 2x² + 5x − 7 + 1/(3x − 2).

55:00 Checking a Polynomial Quotient with the Division Algorithm

The result is verified by multiplying 3x − 2 by 2x² + 5x − 7 and adding remainder 1.
The reconstruction returns the original dividend, 6x³ + 11x² − 31x + 15.
Long division is compared with synthetic division as the lesson's next method.

57:00 Synthetic Division: A Shortcut for Divisors x − k

Synthetic division is presented as a coefficient-only shortcut for a linear divisor with leading coefficient 1.
For x + 2, use k = −2, then bring down, multiply by −2, and add across the coefficient row.
Dividing 2x³ − 3x² + 4x + 5 by x + 2 produces quotient 2x² − 7x + 18 and remainder −31.

1:03:00 Synthetic Division Setup and Coefficient Order

Write k for the divisor x − k, followed by the dividend's coefficients in descending exponent order.
Bring down the leading coefficient, then repeatedly multiply by k and add to the next coefficient.
Interpret the bottom row as quotient coefficients, with the final entry as the remainder.

1:07:00 Synthetic Division of 5x² − 3x − 36 by x − 3

Setting x − 3 = 0 gives k = 3 for the synthetic-division row.
Using coefficients 5, −3, and −36 gives bottom-row values 5, 12, and 0.
The quotient is 5x + 12 and the zero remainder confirms x − 3 is a factor.

1:12:00 Practice: Synthetic Division of a Cubic by x + 2

The practice polynomial is 4x³ + 10x² − 6x − 20, divided by x + 2.
Since x + 2 = 0 gives k = −2, the coefficient row uses −2, 4, 10, −6, and −20.
The class works through the multiply-and-add steps before reviewing the quotient.

1:23:00 Cubic Synthetic-Division Result and Graphical Factor Check

Synthetic division yields 4x² + 2x − 10 with remainder 0.
A zero at x = −2 means x + 2 is a factor of 4x³ + 10x² − 6x − 20.
On a graph, an x-intercept at −2 provides a visual check of that factor.

1:29:00 Cake Dimensions as a Polynomial Volume Problem

A sheet cake has volume 351 cubic inches and is modeled as a rectangular solid.
Its length is 4 inches greater than its width, and its height is one-third of its width.
The setup motivates writing and solving a cubic equation for the cake-pan dimensions.

1:31:00 Remainder Theorem: Why f(k) Is the Remainder

Write f(x) = (x − k)q(x) + r when dividing by x − k.
Because x − k is linear, r is a constant; substituting x = k makes the product term zero.
Therefore, the remainder from dividing f(x) by x − k equals f(k).

1:35:00 Evaluating a Fourth-Degree Polynomial at x = 2

For f(x) = 6x⁴ − x³ − 15x² + 2x − 7, synthetic division by x − 2 uses k = 2.
The synthetic row ends with remainder 25, so f(2) = 25.
Direct substitution into the polynomial confirms the same value.

1:39:00 Remainder-Theorem Practice at x = −3

The final practice evaluates 2x⁵ − 3x⁴ − 9x³ + 8x² + 2 at x = −3.
The missing x-term is represented by a zero coefficient, giving the row 2, −3, −9, 8, 0, 2.
The synthetic-division result is reviewed after students try the problem.

1:46:00 Completing Synthetic Division with a Zero Placeholder

Using k = −3 produces quotient 2x⁴ − 9x³ + 18x² − 46x + 138.
The final synthetic-division entry is −412, so the remainder theorem gives f(−3) = −412.
A zero placeholder preserves the correct coefficient positions whenever a power, such as the x-term, is missing.

Transcript excerpt

0:01 All right, all right, we're good to go. All right? So, let's just go over this how-to one more time before we try an example problem. So, whenever we're given a polynomial function, right? This is how to sketch the graph of it, right? Step one is to find the intercepts, right? So, I'm talking about the x-intercepts and the y-intercepts, right? Step two is to check for symmetry, right? So, if the function is an even function, its graph is symmetrical about the y-axis. If the That is, if f of -x, like if you plug in -x for x, you'll get out f of x. If the function is an odd function, its graph is symmetrical about the origin. That is, if you plug in -x, you're going to get -f of x, right?

0:50 Use the multiplicities of the zeros to determine the behavior of the polynomial at the x-intercepts, right? And then determine the end behavior by examining the leading term, right? And then we're going to use the end behavior and the behavior at the intercepts to sketch the graph, right? To be honest with you, there's no real rule between three, four, and five. You could do that in whatever order like fits you or whatever, you know, whatever you order you like. Um ensure that the number of turning points does not exceed one less than the degree of the polynomial, right? So, let's recap on number six, right? If my polynomial

1:39 is x to the fifth power, I should have at most x to the fourth power terms, right? In the middle of the graph. Right? So, always one less power, the one less terms than the highest power of the polynomial, right? And optionally you can use a calculator to check your graph, right?

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Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Christopher Lee.